1. ## Re: Arc Length Problem

Because you didn't listen to what I was saying. What angle of \displaystyle \begin{align*} \theta \end{align*} makes \displaystyle \begin{align*} \sin{ \left( \theta \right) } = \frac{1}{\sqrt{2}} \end{align*}?

2. ## Re: Arc Length Problem

$\displaystyle \arcsin(\dfrac{1}{\sqrt{2}}) = 45 = \dfrac{\pi}{4}$

45 degrees or $\displaystyle \dfrac{\pi}{4}$, but is that the answer to the whole problem?

Originally Posted by Prove It
Because you didn't listen to what I was saying. What angle of \displaystyle \begin{align*} \theta \end{align*} makes \displaystyle \begin{align*} \sin{ \left( \theta \right) } = \frac{1}{\sqrt{2}} \end{align*}?

3. ## Re: Arc Length Problem

You have been given very good hints and responses here and have NOT done what was suggested.

Again, if $\displaystyle \sqrt{2}sin(\theta)= x= 0$, what is $\displaystyle \theta$? If $\displaystyle \sqrt{2}sin(\theta)= x= 1$ what is $\displaystyle \theta$?

4. ## Re: Arc Length Problem

$\displaystyle \sqrt{2 - x^{2}}$

$\displaystyle 0\leq x \leq 1$

$\displaystyle y = (2 - x^{2})^{2}$

$\displaystyle y' = -x(2 - x^{2})^{-1/2}$

$\displaystyle \int_{0}^{1} \sqrt{1 + (x(2 - x^{2})^{-1/2})^{2}}$

$\displaystyle \int_{0}^{1} \sqrt{1 + \dfrac{x^{2}}{2 - x}}$

Let's stop here, and really look at the next move. Because I still don't really understand this part.

5. ## Re: Arc Length Problem

Originally Posted by Jason76
$\displaystyle \sqrt{2 - x^{2}}$

$\displaystyle 0\leq x \leq 1$

$\displaystyle y = (2 - x^{2})^{2}$

$\displaystyle y' = -x(2 - x^{2})^{-1/2}$

$\displaystyle \int_{0}^{1} \sqrt{1 + (x(2 - x^{2})^{-1/2})^{2}}$

$\displaystyle \int_{0}^{1} \sqrt{1 + \dfrac{x^{2}}{2 - x}}$

Let's stop here, and really look at the next move. Because I still don't really understand this part.
Put it over a common denominator:

$\displaystyle \int_{0}^{1} \sqrt{+ \dfrac{(2 - x) + x^2}{2 - x}} dx$

And don't forget your differential. It is vital that you do not forget your differential.

6. ## Re: Arc Length Problem

Find arc length on interval $\displaystyle 0 \leq x \leq 1$

$\displaystyle \sqrt{2 - x^{2}}$

$\displaystyle 0\leq x \leq 1$

$\displaystyle y = (2 - x^{2})^{2}$

$\displaystyle y' = -x(2 - x^{2})^{-1/2}$

$\displaystyle L = \int_{0}^{1} \sqrt{1 + (x(2 - x^{2})^{-1/2})^{2}} dx$

$\displaystyle L = \int_{0}^{1} \sqrt{1 + \dfrac{x^{2}}{2 - x^{2}}}dx$

Get a common denominator. Eventually you get the next line (after you get the common denominator and add everything up)

$\displaystyle L = \int_{0}^{1} \sqrt{\dfrac{2}{2 - x^{2}}} dx$

Preparing for trig substitution, take the constant out.

$\displaystyle L = \sqrt{2} \int_{0}^{1} \sqrt{\dfrac{1}{2 - x^{2}}} dx$

Now time for trig substitution.

$\displaystyle a = 2$

$\displaystyle x = 2\sin\theta$

$\displaystyle dx = 2\cos\theta$

$\displaystyle 4 - (2\sin\theta)^{2} = 4 - 4\sin^{2}\theta = 4(1 - \sin^{2}\theta) = 4\cos^{2}\theta$

tex]L = \sqrt{2} \int_{0}^{1} \dfrac{1(2\cos\theta)}{\sqrt{4\cos^{2}\theta}}[/tex]

$\displaystyle L = \sqrt{2} \int_{0}^{1} \dfrac{2\cos\theta}{\sqrt{4\cos^{2}\theta}}$

$\displaystyle L = \sqrt{2} \int_{0}^{1} \dfrac{2\cos\theta}{2\cos \theta}$

$\displaystyle L = \int_{0}^{1} \sqrt{2}$ What to do at this point because no trig intervals to work with? Change limits? We looked at this before, but still don't understand.

7. ## Re: Arc Length Problem

Originally Posted by Jason76
Find arc length on interval $\displaystyle 0 \leq x \leq 1$

$\displaystyle \sqrt{2 - x^{2}}$

$\displaystyle 0\leq x \leq 1$

$\displaystyle y = (2 - x^{2})^{2}$

$\displaystyle y' = -x(2 - x^{2})^{-1/2}$

$\displaystyle L = \int_{0}^{1} \sqrt{1 + (x(2 - x^{2})^{-1/2})^{2}} dx$

$\displaystyle L = \int_{0}^{1} \sqrt{1 + \dfrac{x^{2}}{2 - x^{2}}}dx$

Get a common denominator. Eventually you get the next line (after you get the common denominator and add everything up)

$\displaystyle L = \int_{0}^{1} \sqrt{\dfrac{2}{2 - x^{2}}} dx$

Preparing for trig substitution, take the constant out.

$\displaystyle L = \sqrt{2} \int_{0}^{1} \sqrt{\dfrac{1}{2 - x^{2}}} dx$

Now time for trig substitution.

$\displaystyle a = 2$

$\displaystyle x = 2\sin\theta$

$\displaystyle dx = 2\cos\theta$

$\displaystyle 4 - (2\sin\theta)^{2} = 4 - 4\sin^{2}\theta = 4(1 - \sin^{2}\theta) = 4\cos^{2}\theta$

tex]L = \sqrt{2} \int_{0}^{1} \dfrac{1(2\cos\theta)}{\sqrt{4\cos^{2}\theta}}[/tex]

$\displaystyle L = \sqrt{2} \int_{0}^{1} \dfrac{2\cos\theta}{\sqrt{4\cos^{2}\theta}}$

$\displaystyle L = \sqrt{2} \int_{0}^{1} \dfrac{2\cos\theta}{2\cos \theta}$

$\displaystyle L = \int_{0}^{1} \sqrt{2}$ What to do at this point because no trig intervals to work with? Change limits? We looked at this before, but still don't understand.
First note that the first mistake you made was in the line under "Now time for trig substitition". You have:

$\displaystyle L = \sqrt 2 \int_0^1 \dfrac 1{\sqrt{2 - x^2} } dx$

In order to make the substitution you want, you need to see this in the form:

$\displaystyle L = \sqrt 2 \int_0^1 \dfrac 1{\sqrt{a^2 - x^2} } dx$

You have set $\displaystyle a = 2$ whereas instead you needed .... ?

Oh, and be careful with your indices, you've written:
$\displaystyle y = (2 - x^{2})^{2}$
up the top there, which is wrong.

Page 2 of 2 First 12